No, you are not. you need to complete the square, not factor. If you can write it as y= , Then when x= a, y= b and, if so y> b. The vertex is at (a,b).

Your vertex is correct but the equation is wrong. You have the equation for a parabola with horizontal directrix but x= 1 is vertical. Your equation should be .2nd problem:

Write the equation of the parabola with focus (3,4) and directrix x=1.

For the vertex I got (2,4)

So the equation so far is (x-2)^2=4P(y-4)

When y= 5, say, , so x- 2= 1/4P and x= 2+ 1/(4P). For any parabola the distance from a point on the parabola to the focus and to the directrix must be the same. The distance from (2+ 1/(4P), 5) to (3,4) is and the distance to x= 1 is 2+ 1/(4P)- 1= 1+ 1/(4P). Set those equal and solve for P.Help from here getting P would be great also assuming I've done the work correctly...

Thanks a bunch!